I am currently having trouble finding all rows that contain the following data in a table "%2" or "%20".
For Example: 'Susan rides%2bike across town'.
SELECT * FROM [my_table] WHERE [description] LIKE % %2 %
I know this example will not work, however is there a way to search for these values "%2" or "%20" using the SQL "LIKE" operator? As I am required to find and replace them with a white space.
SELECT * FROM my_table WHERE description LIKE '%\%2%' ESCAPE '\';
For the LIKE keyword you can specify the ESCAPE sequence afterwards. In this example the leading backwards slash escapes the % so it will be interpreted as literal value.
Read more about it in the documentation: https://learn.microsoft.com/en-gb/sql/t-sql/language-elements/like-transact-sql?view=sql-server-ver15
you can test it in this SQL Fiddle.
Thank you for this.
SELECT * FROM [my_table] WHERE [description] LIKE '%[%]2%'
The solution for the above stated probelem is
select * from my_table where description like'%[%]2%'
Related
I need to include the special character "%" in my LIKE clause in a SQL query, e.g.:
Select * From Some_Table Where Field_Name Like 'bla%bla&2';
How do I write that?
If you want to match Field_Name values that contain 'bla%bla&2', then you need to write this:
set define off
Select * From Some_Table Where Field_Name Like '%bla\%bla&2%' escape '\';
You get to specify which character you want to use to escape a following character (thanks should go to mathguy, not me). You also have to set define off to prevent sqlplus from trying to substitute values in a string.
If, however, you want to match Field_Name values that exactly equal the given string, then you do this instead:
set define off
Select * From Some_Table Where Field_Name = 'bla%bla&2';
If I am not mistakend you escape them with a backslash (\)
Select * From Some_Table Where Field_Name Like 'bla\%bla&2' ESCAPE '\';
Use escape \ to treat is a literal
SELECT *
FROM Some_Table
WHERE Field_Name LIKE 'blah\%'|| 'blah' ||'&'|| '2';
I'll guess that you're using a tool which treats &n, where n is a digit, as a variable marker. If you're using SQL*Plus or SQL Developer you'd need to issue the SQL*Plus command SET DEFINE OFF. Other tools may use other methods to accomplish this.
Best of luck.
I do not think the backslash escape character will work here for the ampersand. Instead, you will want to divide your search into concatenated strings using double pipes. Use single quotes around each piece of literal text. Next, replace the & with chr(38) which is the ampersand. You can see this with:
select chr(38) from dual
You will still want to include the backslash before the % and finish your statement with escape '\'. Notice, I did not quote the chr(38).
select * From Some_Table Where Field_Name Like 'bla\%bla'||chr(38)||'bla' escape '\'
I am after all cells containing the 'LINE SEPARATOR' (U+2028) unicode point. Normally this is encoded as \u+2028 or something similar. However googling how this translates to SQL has given various options none of which seem to work ((N'2028'), set #hexstring = '2028';, vchar(2028))
SELECT * FROM myTable WHERE desc LIKE '% [SOME WAY TO ESCAPE U+2028 ] %'
ANSI SQL answer, may or may not work with Postgresql.
SELECT * FROM myTable WHERE desc LIKE U&'%\2028%'
Alternative solution using regexp:
select * from mytable where desc ~ '\x2028';
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Possible Duplicate:
SQL Server LIKE containing bracket characters
I am having a problem with pattern matching.I have created two objects say,with codes
1)[blah1]
2)[blah2] respectively
in the search tab,suppose if i give "[blah" as the pattern,its returning all the strings
i.e., [blah1] and [blah2]
The query is
select *
from table1
where code like N'%[blah%'
I guess the problem is with the condition and special characters. Please do revert if you have as solution. Is there any solution where we can escape the character"[". I tried to change the condition as N'%[[blah%'.But even then its returning all the objects that is in the table.
When you don't close the square bracket, the result is not specified.
However, the story is different when you close the bracket, i.e.
select *
from table1
where code like N'%[blah%]%'
In this case, it becomes a match for (any) + any of ('b','l','a','h','%') + (any). For SQL Server, you can escape characters using the ESCAPE clause.
select * from table1 where code like N'%\[blah%\]%' escape '\'
SQL Fiddle with examples
You can escape a literal bracket character this way:
select *
from table1
where code like N'%[[]blah%'
Source: LIKE (Transact-SQL), under the section "Using Wildcard Characters As Literals."
I guess this is Microsoft's way of being consistent, since they use brackets to delimit table and column identifiers too. But the use of brackets is not standard SQL. For that matter, bracket as a metacharacter in LIKE patterns is not standard SQL either, so it's not necessary to escape it at all in other brands of database.
As per My understanding, the symbol '[', there is no effect in query. like if you query with symbol and without symbol it shows same result.
Either you can skip the unwanted character at UI Level.
select * from table1 where code like '%[blah%'
select * from table1 where code like '%blah%'
Both shows same result.
How to write a regular expression to match a string if at least 3 characters from the start are matching?
Here is how my SQL query looks right now -
SELECT * FROM tableName WHERE columnName REGEXP "^[a-zA-Z]{3}someString";
You cannot use CONCAT or alike with REGEX, it will fail. Easiest way to do it, is:
$query = 'SELECT * FROM Test WHERE colb REGEXP "^'.substr($mystring,0,3).'"');
Another is:
SELECT * FROM Test WHERE LEFT(colb, 3) LIKE "{$mystring}%"
Please use jQuery and jqSQL plugin. Note that symbol $ must be escaped in SQL query with this plugin.
I would like to select all records that have an underscore character in their 11th character,
so i try this:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '%%%%%%%%%%_%%%'
but this doesnt work as expected, can someone help?
Just use the "SUBSTRING" function :
SELECT * FROM "BOM_SUB_LEVEL" where SUBSTRING(TOP_CODE, 11, 1) = "_"
Marc
For a single character wildcard use _. For multiple characters wildcards, use %. To escape a "real" appearance of _, use \_ (thanks Bill!).
Try the following code:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '___________\_%'
To further elaborate following Dav's comment, note that '%%%' is exactly the same as '%', since by definition '%' covers multiple characters.
pervasive uses _ to match any single character and \_ to actually match an underscore.
so the select would be:
SELECT * FROM "BOM_SUB_LEVEL" where TOP_CODE like '___________\_%'
LIKE % can mean any number of characters, use LIKE _ to mean just one. Since you're looking for an underscore, you need to escape it with !.
SELECT * FROM BOM_SUB_LEVEL WHERE TOP_CODE LIKE '__________!_%'
The % is not a per character wildcard, its a beginning and end of string wild card.
i.e. if I want to find all rows that have "car" in them, I would do this:
Select * from myTable where myCol LIKE '%car%'
If I wanted just the rows that STARTED with car:
Select * from myTable where myCol LIKE 'car%'
and ended with car:
Select * from myTable where myCol LIKE '%car'
% is a wildcard and can replace an character, or combination of characters. Use ? instead which replaces a single character.
You can try something like: (play with the numbers, I don't have pervasive to test with)
SELECT *
FROM BOM_SUB_LEVEL
where SUBSTRING(TOP_CODE, 11,1) = '-'